Sequences, subsequences and limits

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Homework Statement


Let (rn) be an enumeration of the set Q of all rational numbers. Show that there exists a subsequence (rnk) such that limk\rightarrow\infty rnk = +\infty


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The Attempt at a Solution



Im not sure how to even attack this
 
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If you can make your sequence strictly increasing and unbounded, then it works. Remember that there are only finitely many possible rational numbers before the nth rational number, and yet there are infinitely many rational numbers bigger than whatever the nth rational number is.
 
So how about this: Let (rnk) be a sequence of real numbers such that:

1. rn1 > a, a\epsilon Q
2. (rnk) is strictly increasing
ie. rn1 < rn2 ... < rnk >

Then the lim_{k \rightarrow \infty} r_{n}_{k} = \infty
 
If your 'a' is simply any element of the rational numbers, then your r_n_1 is going to be the supremum of the rationals, which is not in the rationals.

Also, after you show that something's strictly increasing, you have to show that it's also unbounded, meaning that for any M, you can pick r_n_k such that r_n_k > M. After all, the partial sums of sigma from i=1 to infinity of (1/2)^i are constantly increasing, but always less than 2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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